It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?
Answer: B
On 31st December, 2005 it was Saturday.
Number of odd days from 2006 to 2009 = (1 + 1 + 2 + 1) = 5 days.
On 31st December 2009, it was Thursday.
Thus, on 1st Jan, 2010 it is Friday
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What was the week's day on seventeenth June, 1988?
Answer: D
17th June, 1998= (1997 year Period structure 1.1.1998 to 17.6.1998)
Odd days in 1600 year=0.
Odd days in 300 year = (5*3) =1.
97 year has 24 jump year +73 standard years.
= 24*2 + 73*1
= 48 + 73
= 121 odd days
= 17 weeks + 2 odd days = 2 odd days
= 2 odd days.
∴ 1998 years have (0+1+2) = 3 odd days
Jan to June = (31+29+31+30+31)
= 152 days
Add 17 days of June.
= 152 + 17
= 169 days
= 24 weeks + 1 days
= 1 odd day.
∴ Total number of odd day = 3 + 1 = 4 odd days.
Hence 17.06.1998 was Thursday.
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The second day of a month is Sunday, What will be the last day of the next month which has 31 days ?
Answer: D
The 2nd is a Sunday, so the 1st is a Saturday. But the weekday on which the next month begins depends on how long the current month is.
Since the current month is not identified, its length is unknown and the answer cannot be determined.
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On 24th Nov, 2007 Thursday falls. What day of the week was it on 10th Nov, 2006 ?
Answer: A
The year 2006 is an ordinary year. So, it has 1 odd day.
So, the day on 24th Nov, 2007 will be 1 day beyond the day on 24th Nov, 2006.
But, 24th Nov, 2007 is Thursday.
24 - 10 = 14 days.
Therefore, 2 weeks ago it is same day.
Thus, 10th Nov, 2006 is one day before 10th Nov, 2007 i.e. it is Wednesday.
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On 1st Jan, 2005 Saturday falls. What day of the week was it on 1st Jan, 2006?
Answer: C
Only the remainder on division by 7 matters - the odd days.
2005 is not divisible by 4, so it is an ordinary year.
\[ 365 \rightarrow 1 \text{ odd day} \]
\[ \text{Saturday} + 1 = \textbf{Sunday} \]
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The calendar of the year 2006 was used before in the year?
Answer: B
Count odd days backwards from 2006 until they reach a multiple of seven.
\[ 2005:1,\ 2004:2,\ 2003:1,\ 2002:1,\ 2001:1,\ 2000:2,\ 1999:1,\ 1998:1,\ 1997:1,\ 1996:2,\ 1995:1 \]
\[ \text{total} = 14 \quad\Longrightarrow\quad 0 \text{ odd days} \]
So the 2006 calendar had been used in 1995.
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On what dates of April, 2001 did Wednesday fall?
Answer: B
We shall find the day on 1st April, 2001.
1st April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001)
Odd days in 1600 years = 0
Odd days in 400 years = 0
Jan. Feb. March April
(31 + 28 + 31 + 1) = 91 days 0 odd days.
Total number of odd days = (0 + 0 + 0) = 0
On 1st April, 2001 it was Sunday.
In April, 2001 Wednesday falls on 4th, 11th, 18th and 25th.
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Jan.1 2008 is Tuesday. What date of the week lies on Jan 1, 2009?
Answer: C
The year 2008 is a jump year.
So, it has 2 odd days.
First day of the year 2008 is Tuesday (Given).
In this way, first day of the year 2009 is
2 days past Tuesday.
Subsequently, it will be Thursday.
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January 1, 2008 is Tuesday. What day of the week lies on Jan 1, 2009?
Answer: C
Only the remainder on division by 7 matters - the odd days.
2008 is divisible by 4 and is not a century year, so it is a leap year with 366 days.
\[ 366 = 52 \times 7 + 2 \quad\Longrightarrow\quad 2 \text{ odd days} \]
\[ \text{Tuesday} + 2 = \textbf{Thursday} \]
An ordinary year advances the date by one weekday; a leap year by two.
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If Mar 18th,1994 falls on Friday then Feb 25th,1995 falls on which day?
Answer: C
Count the days from one date to the other and reduce modulo 7.
From 18 March 1994 to 18 March 1995 is a full ordinary year, 365 days, so
\[ 365 \equiv 1 \pmod 7 \]
Going back from 18 March 1995 to 25 February 1995 is 21 days, which is exactly 3 weeks and so leaves no odd day. Net shift from the given Friday:
\[ +1 - 0 = +1 \quad\Longrightarrow\quad \text{Friday} + 1 = \textbf{Saturday} \]
1995 is not a leap year and February 1995 has 28 days, so no extra day creeps in.
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